Problem: Factor the following expression: $x^2 - 19x + 90$
When we factor a polynomial, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + a)(x + b) &=& xx &+& xb + ax &+& ab \\ \\ &=& x^2 &+& {(a + b)}x &+& {ab} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + a)(x + b) }&\hphantom{=}&\hphantom{ xx }&\hphantom{+}&\hphantom{ (a + b)x }&\hphantom{+}& \\ &=& x^2 & & {-19}x& +& {90} \end{eqnarray} $ The coefficient on the $x$ term is $-19$ and the constant term is $90$ , so to reverse the steps above, we need to find two numbers that add up to $-19$ and multiply to $90$ You can try out different factors of $90$ to see if you can find two that satisfy both conditions. If you're stuck and can't think of any, you can also rewrite the conditions as a system of equations and try solving for $a$ and $b$ $ {a} + {b} = {-19}$ $ {a} \times {b} = {90}$ The two numbers $-9$ and $-10$ satisfy both conditions: $ {-9} + {-10} = {-19} $ $ {-9} \times {-10} = {90} $ So we can factor the expression as: $(x {-9})(x {-10})$